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\title{单摆的微分方程}
\author{五六七}
%\date{2025年10月8日}

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\section{单摆的微分方程}
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\begin{frame}[allowframebreaks]{单摆的微分方程}

\vspace{-0.3cm}

{\color{red}例子1. 导出单摆方程、求解、画出单摆运动的相图。} 

\begin{center}
\includegraphics [height=0.6\textheight, width=0.6\textwidth]{pic/pendulum-2.png}
\end{center}

设摆线长为 $\ell$, 设小球的质量为 $m$. 设摆线与垂线的有方向的夹角为 $x$. 

根据牛顿第二运动定律，可得 $$m \left(\ell \frac{d^2x}{dt^2} \right) = -mg\sin x.$$

记 $a=\sqrt{\frac{g}{\ell}}$, 则上述微分方程可以写成 $$\frac{d^2x}{dt^2} + a^2\sin x = 0. $$

这是一个自治的二阶微分方程。

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\section{单摆方程的精确解}
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\begin{frame}[allowframebreaks]{单摆方程的精确解 }

\vspace{-0.3cm}

原方程乘以 $\frac{dx}{dt}$ 可得 $$\frac{dx}{dt}\frac{d^2x}{dt^2} + a^2\sin x \frac{dx}{dt} = 0. $$

直接积分可得 $$\frac{1}{2} \left( \frac{dx}{dt} \right)^2 - a^2\cos x = -\frac{1}{2}C_1. $$

开根号得到一阶微分方程 $$\frac{dx}{dt} = \pm \sqrt{2a^2\cos x - C_1}.$$


分离变量可得 $$\frac{dx}{\pm \sqrt{2a^2\cos x - C_1}} = dt. $$

积分可得 $$\int \frac{dx}{\pm \sqrt{2a^2\cos x - C_1}} = t + C_2. $$

设初始位置和初始速度分别为 $x(t_0)=x_0, x'(t_0)=v_0$.  

设单摆的振幅为 $A$, 即 $x(t_1)=A, x'(t_1)=0$. 代入上述第3步，可得 $$C_1=2a^2\cos A. $$


设单摆的振幅为 $A$ 时的周期为 $T$, 则可得 $$\frac{1}{4}T = \int_0^A \frac{dx}{\sqrt{2a^2\cos x - 2a^2\cos A}}. $$

因此单摆的周期与振幅有关，即单摆没有等时性。

设变量代换 $x=Au$, 则可得周期 $$T=\frac{2\sqrt{2}A}{a} \int_0^1 \frac{du}{\sqrt{\cos Au - \cos A}}. $$

当 $A\approx 0$ 时，可得 $T\approx \frac{2\pi}{a}. $



单摆方程的相图

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\includegraphics [height=0.7\textheight, width=0.7\textwidth]{pic/pendulum-phase-diagram.png}
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\section{单摆方程的近似解}
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\begin{frame}[allowframebreaks]{单摆方程的近似解 }

\vspace{-0.3cm}

在振幅很小时，可设 $\sin x \approx x$. 于是微分方程简化为 $$\frac{d^2x}{dt^2} + a^2 x = 0. $$

乘以 $2\frac{dx}{dt}$ 可得 $$2\frac{dx}{dt}\frac{d^2x}{dt^2} + 2a^2 x\frac{dx}{dt} = 0. $$

积分可得 $$\left(\frac{dx}{dt}\right)^2 + a^2x^2 = C_1^2. $$

开根号可得 $$\frac{dx}{dt} =  \pm \sqrt{C_1^2 - a^2x^2}. $$

分离变量可得 $$\frac{dx}{\pm \sqrt{C_1^2 - a^2x^2}} = dt. $$

积分可得 
\begin{eqnarray*}
\int \frac{dx}{\pm \sqrt{C_1^2 - a^2x^2}} = t + C_2. 
\end{eqnarray*}

左边可以初等积分，得到 
\begin{eqnarray*}
\frac{1}{a} \arcsin \frac{ax}{C_1} = t + C_2. 
\end{eqnarray*}

记 $A=\frac{C_1}{a}$, $D=aC_2$, 可得单摆方程的近似解 $$x = A\sin(at+D). $$

可见此时幅角 $x$ 的变化规律是正弦函数。振幅为 $A$. 

在振幅很小时，单摆近似为简谐振动，周期为 $$T=\frac{2\pi}{a}=2\pi\sqrt{\frac{\ell}{g}}.$$ 

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\section{等时摆}
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\begin{frame}[allowframebreaks]{5.1.18. 等时摆 }

\vspace{-0.3cm}

为什么单摆的振动实际上不是等时的？%什么是进动？

{\color{red}问题：什么是等时摆？} %tautochrone

答：加限制的单摆，使得周期与振幅无关。



\begin{figure}[ht]
\includegraphics [height=0.6\textheight, width=0.5\textwidth]{pic/christiaan-huygens.jpg}
\caption{Christiaan Huygens and his pendulum}
\end{figure}


\begin{figure}[ht]
\includegraphics [height=0.6\textheight, width=0.5\textwidth]{pic/cycloidal-pendulum-demonstration.png}
\caption{等时摆}
\end{figure}

% \url{https://www.8sa.net}, \hspace{0.2cm} %/who-is-christiaan-huygens-what-did-christiaan-huygens-discover/}, \\
% \url{https://commons.wikimedia.org}%/wiki/File:Cycloidal_pendulum_demonstration.png}

等时曲线：
\url{https://zhuanlan.zhihu.com/p/379334316}

等时摆：
\url{https://mathworld.wolfram.com/TautochroneProblem.html}

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